Matter & Motion - Winter 2001

Chemistry Homework Answer Key #10

Chapter 12 - Electrochemistry
 

26.  From the table of standard reduction potentials, the reduction of perchlorate is described by the following half-reaction:

ClO4- (aq) + 2 H+ (aq) + 2 e- ® ClO3- (aq) + H2O E°red = +1.19 V

Since Ered is positive, the reduction of perchlorate tends to be favored (it will have a negative DG). Therefore, ClO4- is expected to be a good oxidizing agent, since it will be reduced in the process of oxidizing another species.


43. The half-reactions and corresponding standard potentials for this reaction are:
 

oxidation: 7 H2O (l) + 2 Cr3+ (aq) ® Cr2O72- (aq) + 14 H+ (aq) + 6 e- E°ox = - (1.33 V)

reduction: 2 e- + 2 H+ (aq) + HClO2 (aq) ® HClO (aq) + H2O (l) E°red = 1.64 V

DE°= E°red + E°ox = 1.64 - 1.33 V = 0.31 V

In the overall balanced redox reaction, n = 6.


 


 

log K = 31.45; K = 2.82 ´ 1031
 


 

Given the incredibly large equilibrium constant, we can see that this reaction strongly favors the products. Since the only colored species are Cr2O72-, a product, and Cr3+, a reactant, the expected color of the solution after it reaches equilibrium would be the same as the product, or orange.
 

The only potential problem with this type of qualitative approach is the possibility that Cr3+ is not a limiting reactant. Otherwise, there could be an excess of Cr3+ that would not react, no matter how large K. In this case, 1 mol of Cr3+ was used and would require 1.5 mol of HClO2 to completely react. As it turns out, 2.0 mol of HClO2 are available, so there will be very little Cr3+ remaining at equilibrium.

The solution will be orange.


52. The overall reaction is: Zn (s) + HgO (s) + H2O (l) ® Zn(OH)2 (s) + Hg (l)

Using the hint provided, DG° for the overall reaction is:
 

DG° = [DG°f (Zn(OH)2) + DG°f (Hg)] - [DG°f (Zn) + DG°f (HgO) + DG°f (H2O,l)]
 

DG° = [ -553.3 kJ+ 0 kJ] - [0 kJ -58.56 kJ - 237.18 kJ] = -257.6 kJ
 


 

Note that from the half-reactions, n = 2.
 


 

53. The anode half-reaction is:
Pb (s) + SO42-(aq) --->  PbSO4 (s) + 2 e-

If the mass of lead is 10 kg, the total charge that could passed is:

The maximum work that could be produced is:

w = DG = -nFE = -(9.3 ´ 106 C) (12 V) = -1.1 ´ 108 J

Note that nF is the total charge passed, solved in part a.


65. This reaction occurs in basic solution. The two half-reactions are:

oxidation: Al (s) + 4 OH- (aq) ® Al(OH)4 (aq) + 3 e-

reduction: 2 H2O (l) + 2 e- ® H2 (g) + 2 OH- (aq)

net reaction: 2 Al (s) + 6 H2O (l) + 2 OH- (aq) ® 2 Al(OH)4- (aq) + 3 H2 (g)
 


78. The appropriately written half-reactions are:

 
oxidation: 2I- (aq) ® I2 (s) + 2e-                                             E°ox = -0.535 V

reduction: Cu2+ (aq) + e- ® Cu+ (aq)                                     E°red = 0.158 V

net reaction: 2 Cu2+ (aq) + 2 I-(aq) ® 2 Cu+ (aq) + I2 (s)     DE° = 0.377 V

The sign of DE indicates that this is not a spontaneous process (DG will be positive). So, iodide will not reduce Cu2+ to Cu+.

b) Of course, things change. If the reduction half-reaction above is replaced with the following,

reduction: Cu2+ (aq) + I- (aq) + e- ® CuI (s) E°red = 0.86 V

the overall reaction becomes:

net reaction: 2 Cu2+ (aq) + 4 I- (aq) ® 2 CuI (s) + I2 (s) DE°= 0.325 V

This reaction is spontaneous since the overall potential is positive (DG° will be negative). This is a good illustration of the influence the state of a product can have on the spontaneity of a process. In part a, the copper product was aqueous Cu+; in part b the copper product is solid CuI. These are clearly different and would have different values of DG°f. It shouldn't then be a surprise that the electrochemical potentials would also be different.
 

82. The half reactions are:
 
oxidation: Pb (s) ® Pb2+ (aq) + 2e-                                                         E°ox = + 0.126

reduction: VO2+ (aq) + 2 H3O+ (aq) + e- ® V3+ (aq) + 3 H2O (l)

net reaction: 2 VO2+ (aq) + 4 H3O+ (aq) + Pb (s) ® Pb2+ (aq) + 2 V3+ (aq) + 6 H2O (l)
 


 


 


 


 

E°= 0.463 V
 

DE°= E°ox + E°red
 

E°red = DE°- E°ox = 0.463 -0.126 = 0.337 V
 

b) The equilibrium constant is given by:

K = 4.55 ´ 1015
 


87. At the cathode the reduction of O2 to H2O occurs. The half-reaction is:

reduction: 4 H+ (aq) + O2 (g) + 4 e- ® 2 H2O (l)

b) The value of DE can be obtained from DG for the overall balanced reaction (n = 12), which is:
 

C2H5OH (l) + 3 O2 (g) ® 2 CO2 (g) + 3 H2O (l)

DG° = [3DG°f (H2O, l) + 2DG°f (CO2, g)] - [3DG°f (O2) + DG°f (C2H5OH, l)]

DG° = [ 3 (-237.18 kJ) + 2 (-394.36 kJ)] - [0 kJ - 174.89 kJ] = -1325.4 kJ

DE°= +1.14 V
 
 

c) DE°= E°ox + E°red = 1.14 V = E°ox + 1.229 V
 

E°ox = -0.084 V
 

94. The energy required to produce 1 mol of Al is:
 

This is about 126 times as much energy as that required to melt already refined aluminum (10.7 kJ/mol).